Finding the absolute extrema of $F(x) = 2x + 5\cos(x)$

Question

Find the absolute extrema of $F(x) = 2x + 5\cos(x)$ on the interval $[0,2\pi]$ using the extreme value theorem.

Answer should be 2 ordered pairs.

I got $\arcsin(2/5)$ for the first value of $x$, but can’t figure out the second.

Thanks in advance.

Answer 1

What you are missing is the fact that $\sin(x)=\sin(\pi-x)$ for all $x$.

You found the value $x=\arcsin(\frac25)\approx0.412$. As I mentioned in my comment above, that is a local maximum, and you also need to know about the endpoints of the interval before you can say if it is the absolute max. In this case, we have $F(\arcsin(\frac 25))<F(2\pi)$. The absolute max for the interval is $(2\pi,4\pi+5)$.

From what I said above above, another solution to the equation $\sin(x)=\frac25$ would be $x=\pi-0.412\approx2.73$, and that $x$ value is in the interval $[0,2\pi]$.

The ordered pair for that point, which is the absolute minimum, is approximately $(2.73,0.878)$.

I'd also point out that $\sin(x)=\sin(n\cdot\pi-x)$ for any odd integer value of $n$. You would need this information if you needed to find more extrema on a wider interval.

Answer 2

This function $f(x)= 2x+5\cos{x}$ on $[0,2\pi]$ has one stationary point at $\arcsin{\frac{2}{5}}$. Next you look at the boundary points. It turns out that $f$ has an absolute maximum at the boundary point $x=2\pi$. You should be able to work out the details.

Answer 3

If $f(x)=2x+5\cos x , x \in [0,2\pi], f'(x)=2-5 \sin x, f''(x)=-5 \cos x$ $$f'(x)=0 \implies \sin x=\frac{2}{5} \implies x_1= \sin^{-1} (2/5), x_2=\pi-\sin^{-1} (2/5)$$ $$f''(x_1)<0, f''(x_2)>0 \implies f_{max}=f(\sin^{-1}(2/5))=2\sin^{-1}(2/5)+\sqrt{21},$$ $$ f_{min} =f(x_2)=2[\pi-\sin^{-1}(2/5)]-\sqrt{21}$$ $$f(0)=5, f[2\pi]=4\pi+5$$ So the absolute max is $f(2\pi)=4\pi+5$ and absolute min is $2[\pi-\sin^{-1}(2/5)]-\sqrt{21}$

Answer 4

Derivative vanishes at $\sin^{-1}\dfrac{2}{5}$

plugin and extreme values are

$$ \sqrt{21} + 2(\sin^{-1}\dfrac{2}{5}+ 2 n \pi) $$

Odd multiples are discarded as they give a minimum, can be checked by sign of second derivative. Only alternate roots are considered for maxima.

and the corresponding x-values (first x- value of pair) are

$$ \sin^{-1}(2/5) + 2 n \pi $$

Answer 5

The question says absolute extreme.

$5\cos x \le 5$ and $2x \le 4\pi$ so $2x+ 5\cos x\le 5+4\pi$ so if $2x + 5\cos x$ ever equals $5+4\pi$, which it does at $x = 2\pi$, that will be an absolute (albeit not necessarily a local) maximum. So the absolute maximum is $(2\pi, 5+4\pi)$.

We can't do the same for minimum as $2x \ge 0$ and $5\cos x \ge -5$ but when $2x=0$ we do not have $5\cos x =-5$ for $x =0$ so $2x + 5\cos x > -5$.

For a minimum we can find local minimum via $[2x+5\cos x]'= 2 -5\sin x = 0$ so $\sin x =\frac 25$. So $0<\frac 25 < 1$ there will be one such value $x_1$ in the first quadrant and a second value $x_2$ in the second quadrant. $x_1$ is $\arcsin \frac 25$ and $x_2$ is $\pi-\arcsin \frac 25$.

As $\sin x_1 =\frac 25$ then $\cos x_1 = \sqrt{1 - \frac 4{25}} = \frac {\sqrt{21}}5$. And $x_2$ being in the second quadrant has $\cos x_2 = -\frac {\sqrt {21}}5$.

So the local extrema are $(\arcsin \frac 25, 2\arcsin \frac 25 + \sqrt {21})$ and $(\pi - \arcsin \frac 25, 2\pi - 2\arcsin \frac 25-\sqrt{21})$

Now $\frac 25 < \frac 12$ so $0< \arcsin \frac 25 < \frac \pi 6$ and $5>\sqrt{21} > 4$ so $2\pi -2\arcsin \frac 25-\sqrt{21} < 2\pi - 4 < 7- 4 =3< \sqrt{21}<2\arcsin \frac 25 + \sqrt {21}< 4\pi + 5$

So $(\pi - \arcsin \frac 25, 2\pi - 2\arcsin \frac 25-\sqrt{21})$ is a local minimum and $(\arcsin \frac 25, 2\arcsin \frac 25 + \sqrt {21})$ is a local but NOT an absolute maximum.

Now $2*0 + 5\cos 0 =5> 3 >2\pi -2\arcsin \frac 25-\sqrt{21}$. So the local minimum is the absolute minimum.

So the absolute extrema at the local minimum, $(\pi - \arcsin \frac 25, 2\pi - 2\arcsin \frac 25-\sqrt{21})$ and at the end point, $(2\pi, 4\pi + 5)$.

Answer 6

The possible locations for the extreme values of $f(x)=2x+5\cos x$ on $[0,2\pi]$ are at the endpoints $x=0$ and $x=2\pi$ and the critical points, where $f'(x)=2-5\sin x=0$, which is to say at $x=\arcsin(2/5)\approx0.4115$ and $x=\pi-\arcsin(2/5)\approx2.73$. To determine where the absolute maximum and absolute minimum occur, it suffices to evaluate the function at all four points and see what we get:

$$\begin{align} f(0)&=2\cdot0+5\cos0=5\\ f(\arcsin(2/5))&=2\arcsin(2/5)+5\cos(\arcsin(2/5))\\ &\approx0.823+5\sqrt{1-(2/5)^2}\\ &=0.823+\sqrt{21}\\ &\approx5.406\\ f(\pi-\arcsin(2/5))&=2\pi-2\arcsin(2/5))+5\cos(\pi-\arcsin(2/5))\\ &\approx6.283-.823-5\cos(\arcsin(2/5))\\ &=5.46-\sqrt{21}\\ &\approx0.877\\ f(2\pi)&=2\cdot2\pi+5\cos(2\pi)=4\pi+5\approx17.566\\ \end{align}$$

So the absolute maximum occurs at $x=2\pi$, and the absolute minimum occurs at $x=\pi-\arcsin(2/5)$.

Remark: With a bit of thought, it's unnecessary to explicitly evaluate the function at the four points in order to see where the extrema occur. Since $|\cos x|\le1$ for all $x$, it's clear that $f(x)\le2x+5$, so the absolute maximum necessarily occurs at $x=2\pi$. And since $f''(x)=-5\cos x$ is negative at $x=\arcsin(2/5)$ and positive at $x=\pi-\arcsin(2/5)$, the function has a local maximum at the first critical point and a local minimum at the second. To see that that point is the absolute minimum, it suffices to check that $f(x)\lt f(0)=5$ for some point in $[0,2\pi]$. And $f(\pi/2)=\pi$ does the trick.